求函数y=(sinx-2)/(sinx-1)的值域
问题描述：
求函数y=(sinx-2)/(sinx-1)的值域
答
y=[(sinx-1)-1)]/(sinx-1)
=1-1/(sinx-1)
因为sinx-1∈[-2,0]
所以1/(sinx-1)∈(-∞，-1/2]
- 1/(sinx-1)∈(0,1/2]
y=1-1/(sinx-1)∈(1,3/2]
答
值域：[1,3/2]
∵y=(sinx-2)/(sinx-1)
∴y=(sinx-1-1)/(sinx-1)
=1-1/(sinx-1)
∵sinx∈[-1,1]
∴sinx-1∈[-2,0]
1/(sinx-1)∈[-1/2,0]
- 1/(sinx-1)∈[0,1/2]
1-1/(sinx-1)∈[1,3/2]
应该没出错，若有问题请追问，谢谢~~
答
求函数y=(sinx-2)/(sinx-1)的值域
ysinx-y=sinx-2,故得sinx=(y-2)/(y-1)；
(y-2)/(y-1)≦1.(1)
(y-2)/(y-1)≧-1.(2)
由(1)得(y-2)/(y-1)-1=(-1)/(y-1)≦0,即有1/(y-1)≧0,故得y>1.①
由(2)得(y-2)/(y-1)+1=(2y-3)/(y-1)≧0,故得y